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Can an Oxygen which is double I have a molecule, and part of
it has an Oxygen which is
double bonded to a Carbon, and
no other bonds on the oxygen.
My teacher says it has sp2
hybridisation, but I don't
understand why.
O=C is the bond I am
examining, and I am supposed
to tell what kind of
hybridisation the oxygen
exhibits.
Thanks for the help.
I am asking only about the
hybridisation of the oxygen. I
already know the hybridisation
of the Carbon. Why does the
oxygen have sp2 hybridisation?
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Double bonds involving carbon are sp2 hybridized. When we see a carbon involved in a double bond, its geometry will be trigonal, with 120 degree bond angles, and it will have sp2 hybridization. In a shorthand way trigonal and sp2 are synonyms. Just remember this and you'll do fine. Additional comment: Again, if you are talking about the hybridization of two atoms that are bonded together, the hybridiation of each of those two atoms has to be the same, by definition. Therefore, if the hybridization of a carbonyl carbon is sp2, then so is the hybridization of the carbonyl oxygen... Hope this helps!!! |
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A previous poster said: "Again, if you are talking about the hybridization of two atoms that are bonded together, the hybridiation of each of those two atoms has to be the same, by definition. Therefore, if the hybridization of a carbonyl carbon is sp2, then so is the hybridization of the carbonyl oxygen..." False. And patently nonsensical. Consider propene. Carbon 2 is trigonal, double bond to C1, single bond to C3, single bond to H, sp2 hybridized. It's connected to C3, tetrahedral, single bonds to C2 and three H atoms, so it's sp3 hybridized. Two atoms bonded together with different hybridization. So. Well, draw the Lewis structure, what do you see on the O? A double bond to the C and two lone pairs. Now, VSEPR doesn't tell you explicitly what should happen there, because VSEPR is meant to rationalize the relative position of atoms, not lone pairs you can't see, but an extension of the VSEPR model would suggest that 1 BP + 2 LP would give a trigonal arrangment of electron pair regions, that the "angle" between the LPs would be about 120ยบ. If so, then you'd need sp2 hybrids on O to hold those LPs at those angles, and leave one p orbital available for pi-bonding to C. So you could interpret it that way. Problem. I can offer an alternative, and you have no way to prove it wrong. I can say, no, the O doesn't hybridize. I use pz to make the sigma bond to C, I use px to make the pi bond, I put the two LPs in py and the s orbital. You cannot see the lone pairs, you cannot measure the angle between them, so you cannot know whether my description is better or worse than yours. Can the O be sp2 hybridized? Sure, that's one possible interpretation. Do you know that for certain? No, you don't. More importantly: it doesn't matter. The only reason you invoke hybridization, the ONLY reason, is to rationalize experimentally observed bond angles and bond strengths. Invoking hybridization at O doesn't tell you anything, doesn't explain anything, because there are no angles to observe. You need to invoke it at C, and you need to do it correctly or you predict a result that doesn't happen, but the two offered explanations for hybridization at O (sp2 or none) cannot be distinguished based on the structure of the compound itself. So. Call it sp2 if you like, but if you're feeling mischievous, ask your teacher to prove it, and justify that interpretation, to offer evidence that makes it a better interpretation than that of "no hybridization at O". Because there isn't any. Hybridization is a rationale after the fact, an attempt to explain oberved bond angles -- in the absence of any angles, there's no reason for it. (And, incidentally, the photoelectron spectrum of a ketone, which measures electron energies, better agrees with the no-hybrids at O scheme. The two O LPs have different energies, which is not what sp2 at O predicts.) |
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